MHTCET 20 Apr 2026 Shift 2 Paper

Concept:This is a sequence of reactions involving diazotization, Sandmeyer reaction, reduction, and reaction with nitrous acid.Explanation:Aniline reacts with $\mathrm{NaNO}_{2} / \mathrm{HCl}$ at $0^{\circ} C$ to form benzene diazonium chloride, compound A.$$C_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \xrightarrow[0^{\circ} C]{\mathrm{NaNO}_{2} / \mathrm{HCl}} C_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}$$Compound A undergoes Sandmeyer reaction with $\mathrm{CuCN} / \mathrm{KCN}$, replacing the diazonium group with $-\mathrm{CN}$ to give benzonitrile, compound B.$$C_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl} \xrightarrow{\mathrm{CuCN} / \mathrm{KCN}} C_{6} \mathrm{H}_{5} \mathrm{CN}$$Reduction of benzonitrile (B) with $\mathrm{H}_{2} / \mathrm{Ni}$ gives benzylamine, compound C.$$C_{6} \mathrm{H}_{5} \mathrm{CN} \xrightarrow{\mathrm{H}_{2} / \mathrm{Ni}} C_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}$$Benzylamine (C) is a primary aliphatic amine.When treated with $\mathrm{HNO}_{2}$, it gives benzyl alcohol with the release of nitrogen gas.$$C_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2} \xrightarrow{\mathrm{HNO}_{2}} C_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{N}_{2} + \mathrm{H}_{2} \mathrm{O}$$Thus compound D is benzyl alcohol: $C_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{OH}$.Answer:Option A: $C_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{OH}$